Page 30 - MATINF Nr. 3

30 F. St˘anescu

a+b  a+b   
ñ Å ã 2 ô 0 Z 2 Z b ñ Å ã 2 ô 0 Z x
1 a + b 2 1 a + b 2
2 Z
0
0
= −  − x (f (t)) d t d x +  x − (f (t)) d t d x


2  2 2  2
a x a+b a+b
2 2
a+b a+b
2 Z 2 Z Å ã 2
1 2 2 1 a + b 2
0
0
= (b − a) (f (x)) d x − − x (f (x)) d x
8 2 2
a a
b b
1 Z 2 1 Z Å a + b ã 2 2
0
0
+ (b − a) 2 (f (x)) d x − − x (f (x)) d x.
8 2 2
a+b a+b
2 2
Rezult˘a
b b b
Z Z Z Å ã 2
1 2 0 2 2 1 a + b 0 2
(b − a) (f (x)) d x ≥ (f (x)) d x + x − (f (x)) d x
8 2 2
a a a
é 2
b b
Ñ
Z 1 Z Å a + b ã
2
0
≥ (f (x)) d x + x − f (x) d x
C−B−S 2 (b − a) 2
a a
é 2
Ñ
b b
Z 1 b − a b − a Z
2
= (f (x)) d x + f (b) + f (a) − f (x) d x
2 (b − a) 2 2
a a
b
Z
2 b − a 2
= (f (x)) d x + (f (b) + f (a)) ,
8
a
de unde
b b
Z 1 8 Z
2
0
2
2
(f (x)) d x − (f (a) + f (b)) ≥ (f (x)) d x.
b − a (b − a) 2
a a
Aplicat , ia 5. Fie f : [0, 1] → R o funct , ie continu˘a. Demonstrat , i c˘a:
1 Ñ 1 é 2 1 1
1 Z Z Z Z
2
f (x) d x + 2 f (x) d x ≥ 3 f (x) d x · xf (x) d x.
4
0 0 0 0
Solut ,ie. Fie a ∈ R astfel ˆıncˆat
2
é
1 1 1 1
Ñ
Z Z Z Z
1 2
f (x) d x + (2 + 3a) f (x) d x ≥ 3 f (x) d x · (x + a) f (x) d x.
4
0 0 0 0
1 1 1
2
2
R R 1 R
Dac˘a f (x) d x = t, atunci (2 + 3a) t − 3t (x + a) f (x) d x + f (x) d x ≥ 0 ⇒
4
0 0 0
2
Ñ é
1 1
Z Z
2
∆ t = 9 (x + a) f (x) d x − (2 + 3a) f (x) d x
0 0